A. The main purpose of spectroscopy in organic chemistry is to
determine the structure of a compound.
1. Types of spectroscopy include [but are not limited
to]a) Infrared Spectroscopy, IR.
b) Nuclear Magnetic Resonance, NMR.
c) Mass Spectrometry.
d) UV-Visible Spectroscopy.
2. Other types of spectroscopy have been developed,
such as esr.

II. Infrared Spectroscopy
A. Theory of IR
1. Molecules can vibrate in many ways.
2. The energy required to excite these vibrations
corresponds to electromagnetic radiations in the
infrared region.
3. Different types of bonds absorb specific
frequencies of IR radiation. a) The frequencies are often measured
in wavenumbers (cm-1), the number of waves which would
fit in a 1 cm line.
b) For example, the C=O bond absorbs
energy in the 1660-1770 cm-1 range while the C-O bond
absorbs between 1050-1300 cm-1.
4. The information IR gives the chemist is the
presence (or lack of) specific functional groups in the
compound being analyzed.
5. IR cannot tell you how many atoms of each
type are in the compound.

B. Components of the IR Spectrometer 1. A simple IR spectrometer consists of an IR
energy source which sends energy through a prism.
a) The prism separates the energy into
specific frequencies, just like it separates visible light into
the
colors of the rainbow.
i) In the newer Fourier
Transform IR instruments, prisms are not used; an improved
technology is used instead.
b) The individual frequencies are sent
through the sample cell which holds the compound being
studied.
i) The sample cell can hold
gases, liquids, or solids.
ii) Gases and liquids can be
held in the sample cell as the pure compound ("neat").
iii) Solids can be mixed with
solid KBr (invisible to IR), ground into a fine powder, and
then subjected to pressure
to make a transparent window.
iv) Solids can also be ground
with mineral oil to make a “Nujol mull”
c) When energy is absorbed due to the
presence of a specific bond type vibrating, a detector notes
the
sudden decrease of energy
reaching it (a reference beam bypassing the sample cell is used for
comparison).
d) The detector sends the information to a
recording device.
a) Energy absorptions are seen
as "absoprtion bands" on the IR spectrum.

C. Interpretation of IR Spectra 1. Not every absorption band in an IR spectrum is
significant since certain frequency ranges of IR are
absorbed by several different types of
bonds.
a) Only certain regions of the IR spectra
are meaningful.
2. Some of the important regions of the IR spectrum
to look at include:
a) 3200-3600 cm-1: O-H from
alcohols.
b) 3000-3100 cm-1: C-H from
sp2 C.
c) 2850-3000 cm-1: C-H from
sp3 C.
d) 1660-1770 cm-1: C=O
e) 1600 cm-1: C=C from benzene.
f) 1050-1300 cm-1: C-O

II.
Nuclear Magnetic Resonance
A. Theory
1. The nuclei of some atoms behave as though they
spin on their axis.
a) Electrons have the same behavior.
2. The 1H nucleus (a proton) has two
nuclear spin states with quantum numbers +1/2 and -1/2.
3. Spinning charged particles create a magnetic
field, so the proton has a magnetic moment.
a) The proton acts like a small bar
magnet.
4. The two spin states are identical in energy in the
absence of an applied magnetic field.
a) In absence of an applied magnetic
field, the nuclear magnetic moments are randomly oriented.
5. When placed in a magnetic field the magnetic
moments align either with or against the field.
a) The two spin states are no longer of
equal energy.
i) The +1/2 spin state aligns
with the applied magnetic field and is of lower energy.
ii) The −1/2
spin state aligns against the applied magnetic field and is of higher energy.
b) Example: think of the Earth's magnetic
field as an applied field in which a compass is placed.
i) The compass needle aligns
with the magnetic field of the Earth.
ii) This is of lower energy
than if you were to align the needle in the opposite direction.
6. The nucleus in the lower (+1/2) spin state can be
elevated to the higher-energy spin state by the
absorption of energy (electromagnetic
radiation) equal to the energy difference of the two spin states.
a) This change from the +1/2 to −1/2 spin state is referred to as "spin
flipping".
b) The amount of energy required for this
transition is dependent upon the magnitude of the applied
magnetic field.
c) The greater the applied field, the
higher the required energy to flip the spin.
7. Many nuclei lack the property of spin.
a) Nuclei having an even number of
protons and an even number of neutrons lack spin.
i) Pairs of protons and pairs
of neutrons in such nuclei have opposed spins which cancel each
other out, leaving the
nucleus with no net spin.

B. NMR Spectroscopy 1. The NMR spectrometer consists of a powerful
magnet.
2. A sample is placed in the magnetic field of the
spectrometer.
3. A radio-frequency source irradiates energy through
the sample until the frequency of the energy source
is equal to the energy difference of the
two spin states.
a) At this time the two are said to be in
resonance with each other.
4. The absorption of energy is detected by a
radio-frequency receiver and recorded as a peak on an NMR
spectrum.
5. The fact that radio waves are used to cause the spin
flip demonstrates that the energy difference
between the two spin states is of very low
energy.
a) In a very strong magnetic field of
14,100 gauss, the energy difference for the two spin states of a
proton is only 2.5 x 10-5
kJ/mole.
6. NMR instruments can operate either with a constant
magnetic field and changing the frequency, or
with constant frequency and changing
magnetic field.
a) Either way, the spectra are identical.
7. Not long ago 60 MHz instruments were in common
usage, then 100 MHz, 220 MHz, and 360 MHz
instruments were in use; now 400, 500 and
600 MHz instruments are in use.

C. Chemical Shift 1. If all hydrogen nuclei spin flipped at the same
frequency, NMR spectroscopy would have no value in
compound structure determination since all
NMR spectra would look the same: a single peak at the
same place.

2. Fortunately, the chemical environment of
hydrogen atoms affects the frequency of the energy
absorbed for resonance to take place. 3. In a constant frequency instrument, where the
magnetic field is slowly changed to cause the spin flip, it
is observed that the applied magnetic field
required to spin flip a proton in an organic compound is
greater than that required for a bare
proton.
a) This is because the electrons in a
molecule "shield" the nucleus from the applied magnetic field.
i) This is referred to as
"diamagnetic shielding".
b) This shielding occurs because the
application of the magnetic field results in the electrons moving
and setting up their own
magnetic field which opposes the applied field.
c) The nucleus of the bound proton "feels"
less magnetic field than is actually applied.
d) In order to achieve the spin flip, a
greater magnetic field must be applied as compared to the field
required to flip a bare proton.
4. The change in the resonance position of a nucleus
(due to the chemical environment of the atom) is
referred to as a chemical shift.
5. A peak at a higher field is said to be "upfield".
a) A peak at a lower field is said to be
"downfield".

6. The chemical shift is measured relative to a
substance standard
a) The standard used is TMS,
tetramethylsilane.
b) The NMR signal from TMS is set at 0 (on
the delta scale) since protons in TMS are highly
shielded due to the low
electronegativity of Si.
i) There is therefore a
relatively high electron density around the hydrogen atoms.
ii) Shielding in TMS is greater
than that of most organic compounds, making it an ideal standard.
iii) The spectrum is adjusted
electronically on the chart so that the TMS signal is set to 0.
7. The placement of an NMR signal relative to TMS
varies with magnetic field strength.
a) Because NMR instruments use many
different magnetic field strengths, a unitless measure of
chemical shift independent of
field strength has been developed.
b) The delta scale is the ratio of the
chemical shift of a given signal to the total radio frequency used:
Chemical shift (d)
= (position of signal - position of TMS peak) x 106/radio frequency of spectrometer
c) The values in the above equation are in
Hz.
d) Because the difference in the two
positions is so small, it is multiplied by 106.
i) The delta scale is therefore
expressed in ppm.

8. The chemical shift yields important information.
a) By the magnitude of the shift, some
details of chemical structure can be deduced.
i) The text has a more complete
listing of chemical shifts; below are some representative values.

9. The chemical shift for benzene is around 7 ppm and
that for the protons of alkenes is around 5.
a) Resonance occurs far downfield than
might be expected for hydrocarbons.
b) The effect is due to anisotropy.
i) The pi cloud of electrons of
these systems circulate due to the induced magnetic field,
creating their own
magnetic field.
c) In the case of benzene, the induced
magnetic field opposes the applied field in the middle of the
ring, but is with the applied
field outside the ring.
i) The protons of benzene are
in the space where the secondary magnetic field generated by
the pi electrons of
the pi cloud circulating deshields the protons.
ii) A strong clue to the
presence of the benzene ring in a compound is a NMR signal at ~7 ppm.

d) In the case of alkenes, the induced
field opposes the applied field in the middle of the double
bond, but are with the applied
field in the area of the vinyl hydrogens.

D. Splitting Patterns

1. The NMR spectrum of ethane is a single peak since
there is only one chemical environment for H
2. The NMR spectrum of 1,1,2-trichloroethane should
show how many NMR signals, and what would
you expect their approximate chemical
shifts?

Answer: Two.

The H at C-1: ~6
ppm. It is further downfield than the H’s at C-2 since two Cl are attached to
its C.The H's at C-2: ~4
ppm

a) Two single peaks are NOT observed for
the above compound.
b) A triplet is observed at 5.8 ppm
and a doublet is observed at 3.9 ppm.

3. Multiplets are observed in place of singlets
on the NMR spectrum when hydrogen atoms exist on
adjacent carbon atoms.
a) Multiplets arise from "spin-spin
splitting".
b) The chemical shift is in the center of
the multiplet.
4. Splitting occurs because the magnetic field of a
proton of a hydrogen atom on carbon 1 adds or
subtracts to the applied field felt by the
hydrogen atom on carbon 2.
a) The "effective field" felt by a proton
can be greater, less than, or equal to the applied field,
depending on the orientation of
spin on neighboring hydrogens.
5. The two hydrogens on C-2 of 1,1,2-trichloroethane
experience two different effective magnetic fields
depending upon if the hydrogen on C-1 is in
the spin up or spin down state.
6. The H at C-1 is near two hydrogens on C-2 and as a
result can feel 3 different effective fields.
a) Both C-2 H-spins can be up, both could
be down, or one could be up while the other is down.

b) Notice that the central peak of the
triplet is twice the size of the other two since there are twice
as
many spin combinations possible
than for the other two.
c) The hydrogens on the adjacent carbons
are said to be "coupled".
7. In simple cases, the observed multiplet is equal to
the number of H on adjacent carbon atoms + 1.
This is called the "(n+1) rule"
8. The splitting pattern for the ethyl group is quite
distinctive.
a) Predict what the proton NMR spectrum
would look like for CH­3CH2F, and show the spin
combinations and net spins below
each signal.

9. The following chart shows the type of multiplet
observed as a function of adjacent hydrogens, and the
area ratios of the peaks of the multiplet.
You may recognize this pattern as Pascal’s Triangle

# of
Adjacent[vicinal]Equivalent H's

Number of Peaks

Peak
Descriptionlearn
the lingo!

Area
Ratios

0

1

singlet

1

1

2

doublet

1:1

2

3

triplet

1:2:1

3

4

quartet

1:3:3:1

4

5

quintet

1:4:6:4:1

5

6

sextet

1:5:10:10:5:1

6

7

septet

1:6:15:20:15:6:1

10. Protons of the hydroxyl group on alcohols usually
appear as a singlet and are not split due to rapid
intermolecular exchange of the hydroxyl
protons. The same is true for amine hydrogens.

E. Coupling Constants

1. The distance between peaks in a multiplet, measured
in Hz, is referred to as the coupling constant, J.
2. The coupling constant is independent of the
magnetic field strength of the NMR instrument.
3. For the example fill in the blanks:
The peak for Ha appears as a
doublet. The peak for Hc appears as
a doublet.
a) Because the coupling constants Jab
and Jbc are not equal (Jab = 3.6 Hz while
Jbc = 6.8 Hz) the peak
for Hbis not a
quintet but instead a doubled quartet (or quadrupled doublet.)
i) This makes the multiplet
rather messy rather than being a nice and easy-to-decipher peaks.

4. Unequal coupling constants can result in complex
spectra which are more difficult to interpret.
5. Coupling constants can yield important information
about conformations in cyclohexane rings.
a) The coupling constant for adjacent axial
hydrogens in cyclohexanes is 10-13 Hz.
b) The coupling constant for adjacent
equatorial hydrogens in cyclohexanes is 3-5 Hz.

Jaa'
= 11.1 HzJae = 4.3 Hz

Jee'
= 2.7 Hz

Jae
= 3.0 Hz

6. Coupling constant values can be used for the
determination of stereostructure in alkenes.
a) J is always greater for trans
H’s than for cis H’s for a pair of isomeric cis and trans
alkenes.
i) Jtrans =
11-19 Hz.
ii) Jcis =
5-14 Hz.
b) Notice there is overlap in the J
values.
i) If the vicinal coupling
constant is < 10, one can infer the alkene is the cis isomer.
ii) If the vicinal coupling
constant is > 14, one can infer the alkene is the trans isomer.
iii) If the J value is
between 10 and 14, then the spectra of both isomers must be available
to make a conclusion
(Jcis is always less than Jtrans for
isomeric alkenes.)
c) The NMR spectrum of (E)- and (Z)-3-chloropropenenitrile
illustrate how the coupling constants
can be used to assign
structure.

F.
Integration

1. In proton NMR the area of the peaks for each proton
environment is proportional to the number of
protons in that environment.
2. When the NMR spectrometer is set to the integration
mode it draws a step-like line whose height is
directly proportional to the number of
protons responsible for the signal.
3. Integrations do not actually indicate the precise
number of hydrogens at each chemical environment.
a) They instead yield a ratio of
protons at each environment.
4. The example below illustrates how integration yields
a ratio of protons at each chemical environment.

a) In the above example notice how the
integral ratios are calculated.
5. The information given by the chemical shifts,
splitting patterns, coupling constants, and the
integration can often lead to the structure
of the compound in question.
a) In addition to this we also have
information given by IR (functional groups present), mass
spectrometry (molecular weight
and other information), elemental analysis, and
other spectroscopic methods.

G. C-13 NMR (CMR) 1. NMR experiments may be done with any atom
which has a nucleus with a magnetic spin.
2. Examples of various isotopes with nuclear magnetic
spin states are listed below.
a) The chart shows the frequency and
magnetic field strength for which the nuclei have their
resonance. Note that for a
given isotope, resonant frequency is proportional to applied field.

isotope à

1H

2H

13C

19F

Ho,
gauss

10,000

14,100

42,276

51,480

10,000

10,000

42,276

10,000

42,276

Resonant
Frequency
MHz

42.6

60.0

180

220.0

6.5

10.7

45.3

40.0

169.2

3. A significant difference in proton NMR vs CMR is
that 1H is the most abundant isotope of hydrogen.
a) In CMR the lesser abundant isotope is
observed. About 1% of all carbon atoms are C-13.
b) The lower abundance means a larger
sample must be used in the instrument.

Proton NMR requires about .1 mg
of sample. CMR requires about 1-5 mg of sample.
4. The 1H NMR spectrum ranges typically from
0-12 ppm; the CMR spectrum ranges from 0-210 ppm.
a) Even slight differences in a carbon's
chemical environment results in a significant chemical
shift so that the number of
carbon atoms (of different chemical environment) in the compound
can be determined simply by
adding up the number of signals in the spectrum.
5. A great advantage of CMR is that the NMR experiment
can be run in two different modes.
a) The "off-resonance decoupling"
experiment results in the splitting of each signal for each carbon
atom in the molecule by the
hydrogen atoms attached to it.
i) This means that the number
of hydrogen atoms attached to the carbon atom can be quickly
deduced simply by
counting the peaks of the multiplet and subtracting 1.

b) In the proton-decoupled mode each
carbon nucleus appears as a singlet.
i) The protons are kept from
coupling with the C-13 nuclei by irradiating the sample with
radio waves which
flip the protons: they do not spend enough time in either spin state to
couple with the C-13
nuclei.

ii) Adjacent carbon atoms do
not split the signal in C-13 NMR. Why not?Answer: Since there are so few C-13 atoms (~1%),
the chances of having two C-13
atoms next to each
other in a molecule are small.

6. In CMR as in NMR, electronegative atoms attached to
the carbon atom results in a downfield shift.
a) In addition, replacement of a hydrogen
atom on a carbon with an alkyl group results in a
downfield shift.

Chemical Shift, ppm

10.2 ppm

21.8

30.9

69.0

Answer

C-4

C-1

C-3

C-2

7. In the CMR spectrum of 2-butanol above, assign which
carbons are responsible for each observed
peak.
8. Coupling constants could be used in proton NMR to
distinguish cis and trans isomers of alkenes.
a) The chemical shift of the allylic
carbons in CMR are lower for the cis isomer than the trans
isomer by about 5 ppm, so this
is another way of deciding upon which isomer is being studied.
b) Notice the chemical shifts for the
isomers cis- and trans-2-butene. cis:
11.4 ppm trans:
16.8 ppm

9. Unlike proton NMR, CMR integrals do not yield ratios
of nonequivalent carbons since the peak
intensities in CMR are subject to
distortion.
10. Below are shown some CMR chemical shift
correlations (ppm).

H. Spectral Interpretation 1. It is not necessary to memorize the various
chemical shifts for the different types of chemical
environments.
a) Tables can be consulted.
b) It is helpful to remember some
generalities such as, thinking of the proton NMR spectrum as a
football field:
i) Alkanes resonate around the
10-20 yard line on the right side.
ii) Hydrogens bonded to carbons
in turn bonded to highly electronegative groups (OH, Cl, Br)
resonate near
mid-field.
iii) Inductive shielding
effects are cumulative, so the greater the number of Cl's, O's, etc
bonded to a carbon,
the more the C-H resonance will move downfield (past mid-field to
the other end zone.)
c) In CMR remember that double-bonded
carbons resonate far downfield, C-O and C-X
(X = halogens) moderately far
downfield, and C=O very far downfield.
2. From CMR one can find the number of nonequivalent
carbons and in the off-resonance mode the
number of hydrogens on each carbon.
3. From proton NMR probably the most useful aspect is
the spin-spin splitting patterns.
4. The integrations in proton NMR yield proton ratios
in each chemical environment and should be
evaluated.
5. The chemical shift of ~7 is usually indicative of
the presence of the benzene ring.
a) A signal between 9.5 and 10 usually
indicates the presence of an aldehyde
b) A signal between 10 and 12 usually
indicates the presence of a carboxylic acid.
I. Sample Problems
1. 3-hexanol is treated with boiling sulfuric acid
forming a mixture of isomeric alkenes:

The
mixture was then separated by gas chromatography and the CMR spectra taken
of each fraction.
The following CMR information resulted (in ppm):

Isomer

d,
ppm

A

12.3,
13.5, 23.0, 29.3, 123.7, 130.6

B

13.4,
17.5, 23.1, 35.1, 124.7, 131.5

C

14.3,
20.6, 131.0

D

13.9,
25.8, 131.2

Identify each isomer.

Answer. Isomers
A and B have six signals and therefore must have six nonequivalent carbons (III
and IV). Isomer B has two chemical shifts about 5 ppm greater than Isomer A
(17.5 vs 13.5, 35.1 vs 29.3), which indicates isomer B must be compound III (the
trans isomer) and A must be compound IV.

By
similar analysis, Isomer C must be compound I (the cis isomer) and Isomer D must
be compound II.

2. A 1H NMR experiment is done on a
compound of formula C4H7Cl3. Use the following
proton NMR data to determine the structure of the compound. (s = singlet, d =
doublet; t = triplet; m = multiplet).

Answer.The triplet at .9 (upfield) with an integration of 3 (3
H's) is a methyl group next to a methylene group. CH3CH2-The doublet at 5.8
is so downfield that it is most likely -CHCl2. Being a doublet there
must be only one hydrogen on the adjacent carbon.
The multiplet at 4.3 with an integration of 1 could
account for this adjacent carbon with 1 hydrogen atom; the fact
that it is downfield can be explained by the presence of a Cl atom and
being next to a carbon with two Cl atoms.

-CHCl-

The structure of the compound is likely to be
1,1,3-trichlorobutane

3. A compound of formula C3H7NO­2
(an -NO2 group is present – thank IR for that) has the
following proton NMR spectrum.
Determine the structure of this compound.

Answer: CH3CH2CH2NO2

Ha: 1.0 ppm (t); Hb: 2.0 ppm (m); Hc: 4.4 ppm (t)

4. A compound has the formula C4H7O2Br.
The signal at 10.97 ppm was moved onto the chart since the
chart paper only runs from 0-10 ppm. Determine the
structure based on the proton NMR spectrum.

Answer. CH­3CH2CHBrCO2H

Ha: 1.1 ppm
(t); Hb: 2.0 ppm (m); Hc: 4.2 ppm (t); Hd: 10.97 ppm (s)

Mass SpectrometryDue to limited time, we'll do a flyby of
the "base peak" rather than an Everest-type expedition

Mass Spectrometry for
Chromatographers by James K. Hardy at the University of Akron
http://ull.chemistry.uakron.edu/gcms/
the output from a gas chromatograph [GC] can be analyzed by mass spec [MS]
-- hence, GCMS

III. Mass Spectrometry

A. Theory
1. Molecules are bombarded with a beam of high energy
electrons.
2. The molecules are ionized and fragmented into many
smaller pieces.
3. Most of the particles carry a +1 charge.
a) This means that their mass/charge
ratio, the m/e value, is the mass of the ion.
b) For the case of the molecular cation,
the m/e value is the molecular weight of the compound.
i) This is referred to as the
M+ peak, and M+ is the molecular ion or parent
ion.
4. The different ions produced is sensed by a detector,
which sends the information to a recorder which
plots the signals in relationship to their
intensities.
a) The largest signal is referred to
as the base peak and is given a relative intensity of 100%.
i) The M+ is not
necessarily the base peak.
ii) The M+ peak can
sometimes be extremely small.
b) The intensities of the other peaks
are then expressed relative to the base peak.
c) The mass spectrum is highly
characteristic of a given compound and can be used to prove the
identity of a compound or used
to determine the structure of a compound.
5. The mass spectrum helps in structure determination
by
a) giving a molecular weight for the
compound.
b) indicating certain structural units
existing in the molecule, and
c) sometimes can give a molecular
formula for the compound.

B. The M + 1 Peak 1. The M+ peak is not the peak of
highest m/e value.
2. This is because of the existence of isotopes.
3. Since about 1.1% of all carbon atoms is C-13, some
of the molecular ions have a m/e value 1 unit
higher than the parent peak.
a) The (M + 1) peak can also be due to the
presence of deuterium, although the % abundance of 2H
is only 0.015%.
4. In the case of benzene, it would be expected that
the (M + 1) peak would be about 6.6% as intense as
the M+ peak.
a) Since there are six carbons in benzene,
the % intensity expected is 6 x (1.1%).
5. In compounds containing chlorine a significant (M +
2) peak would be observed since the
% abundance of 37Cl (2 amu's
more massive than the more abundant 35Cl) is about 24.23%.
6. In compounds containing Br, an (M + 2) peak of about
equal intensity as the parent peak is observed
since the two isotopes of Br, 79Br
and 81Br, exist in a 100:98 [~1:1] ratio.
a) “Twin” peaks of about equal intensity
but two units different in mass is a good indication of Br in
the compound.

C. Example

1. Neopentane has been analyzed by mass spectrometry
and found to break into several fragments
including the following.
a) Notice that the fragments formed are not
obvious fragmentation products of neopentane.
i) Rearrangements obviously
must occur.
ii) Stability of carbocations
is a driving force for rearrangements in mass spectrometry.
b) Only about 10-6 gram of
sample is needed for the experiment.

2. Fragmentation patterns and types of
rearrangements are well-known and are a key to the full
utilization of mass spectrometry as a tool
for structure determination.

Ultraviolet and Visible SpectroscopyWe'll take this on in more detail when we do
chapters 10 & 11

A. Theory 1. Ultraviolet light is of much higher energy than
that used in IR and NMR.
2. UV spectroscopy uses electromagnetic radiation in
the range from 200 to 400 nm.
3. Visible spectroscopy uses light in the 400 to 800 nm
range.
4. Both are used in the investigation of the
electronic structures of unsaturated compounds and
measuring the extent of conjugation.
5. The absorbance A of a sample is proportional to its
concentration in solution and the path length of
the sample cell.
a) To correct for the above factors, the
absorbance is converted to molar absorptivity, e.
[formerly called the “molar
extinction coefficient”]

A =
absorbance
c = concentration in mol/Ll
= path length in cm

b) When molar absorptivity is measured at
lmax,
it is cited as emax.
i) It is usually expressed
without units.
6. Solvents used in UV-Visible spectroscopy are
ethanol, methanol, and cyclohexane.
a) These solvents do not absorb in the
uv/visible range.
b) Solvents have an effect on the
wavelength absorbed and the molar absorptivity, so they should be
listed in electronic
spectra data.
7. Absorption of the energy occurs by the excitation of
electrons to higher energy levels.
a) The electrons are promoted from a filled
bonding molecular orbital (p) or a nonbonding
molecular orbital (n) to
an antibonding molecular orbital (p*).
b) These transitions are written as:
i) pàp*
ii) nàp*
c) The compound has been changed from its
ground electronic state to an excited electronic state.
8. The excitations are recorded as electronic
spectra.
9. Sigma bonds require too great an energy to be used
in this technique.

B. Electronic Spectra and Delocalization

1. Electronic spectra can indicate the size and degree
of delocalization in an extended pi system.
2. The greater the number of double bonds conjugated
together in a molecule, the lower the energy of
excitation will be.
a) There will also be a greater number of
peaks in the spectrum.
3. Examples of some molecules and their l max follow.

Compound

l
max
(nm)

ethene

171

1,4-pentadiene

178

1,3-butadiene

217

1,3-cyclohexadiene

259

trans-1,3,5-hexatriene

268

trans, trans-1,3,5,7-
octatetraene

330

4. Notice that the first two compounds above are
not conjugated and have a much higher energy (shorter
wavelength) than the conjugated compounds.
5. Beta-carotene has 11 double bonds in conjugation.
a) The wavelength absorbed is so long that
it is in the visible range, giving beta-carotene its
characteristic orange color.
6. Larger conjugated pi systems have lower-energy
excited states because with more p-orbitals
combining to form molecular orbitals, the
energy gap between bonding and antibonding molecular
orbitals gets smaller.
a) Also, more bonding and antibonding
orbitals are available to rise to more electronic transitions.
7. The structural unit associated with the electronic
transition is referred to as a chromophore.
8. In alkenes and polyenes the electronic transitions
occur from the HOMO (highest occupied molecular
orbital, a pi molecular orbital) to the LUMO
(lowest unoccupied molecular orbital), the pi
antibonding orbital.
a) In carbonyl compounds, it is the nàp*
transition of the C=O group which occurs.
i) Here the electron being
promoted is a nonbonding electron from the lone pair on oxygen.
9. UV-visible spectroscopy does not offer as much info
as NMR or IR, but is still an important tool.

Many thanks to Rod Oka of
MPC for generously sharing his "Lecture Companion" outline,
reproduced here
in extensively modified form by permission, with
web references and other goodies added by me.
Structures drawn using MDL IsisDraw™, ACDLabs ChemSketch™, and
CS ChemOffice ChemDraw™.
Other graphics drawn with MS Excel™ [my favorite program]